Dunno, I don't do step (I'm a bio natsci lmfao) , but judging from my instincts I think that'll be at least 812 marks off because that seemed to be the majority of the work (had to spot x^2=0 approximation and binomial approximation, trigonometric manipulation etc)(Original post by matthewdjones)
How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?
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STEP 2016 Solutions
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 24062016 23:41
Last edited by LTyrosine; 24062016 at 23:45. 
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 24062016 23:51
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y^{2} = 4ax in parametric form is: x = at^{2}, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t^{ 1}.
So the gradient of the normal to the curve at the point with parameter t is t.
Hence, the normal to the curve at Q has gradient q, so has equation
y  2aq = q (x  aq^{2}).
Since this line passes through P, we substitute y = 2ap and x = ap^{2}. Dividing through by a, we have
2 (p  q) = q (p^{2}  q^{2}) = q (p + q) (p  q), so 2 = q (p + q) and hence q^{2} + qp + 2 = 0.
Similarly, r^{2} + rp + 2 = 0.
(ii) Line QR has gradient (2aq  2ar) / (aq^{2}  ar^{2}) = 2(q  r) / [ (q  r) (q + p) ] = 2 (p + q)^{ 1}.
r^{2} + rp + 2 = 0 = q^{2} + qp + 2
=> r^{2}  q^{2} = (r  q) (r + q) = p (q  r)
=> r + q = p
=> gradient of line QR is 2 p^{ 1}
=> equation of QR is y = 2 p^{ 1} (x  ar^{2}) + 2ar.
Let y = 0. Then x = a (r^{2} + rp) = 2a.
So QR passes through ( 2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p^{ 1} x.
So, at T, where OP and QR intersect,
2 p^{ 1} x = 2 p^{ 1} (x  ar^{2}) + 2ar, which implies that
x = 1/2 a (r^{2} + 2rp) = 1/2 a (2) = a.
So T lies on the line x = a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = a into the equation of OP, we find that T has y coordinate 2ap^{ 1}. So the distance of the xaxis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u^{2} + up + 2 = 0, this quadratic must have positive discriminant, i.e. p^{2} > 8, so the modulus of p is greater than 2 (2)^{1/2}.
Hence the modulus of 2ap^{1} is less than a / 2^{1/2}.
^{ }Last edited by riquix; 25062016 at 00:03. 
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 24062016 23:53
(Original post by riquix)
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y^{2} = 4ax in parametric form is: x = at^{2}, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t^{ 1}.
So the gradient of the normal to the curve at the point with parameter t is t.
Hence, the normal to the curve at Q has gradient q, so has equation
y  2aq = q (x  aq^{2}).
Since this line passes through P, we substitute y = 2ap and x = ap^{2}. Dividing through by a, we have
2 (p  q) = q (p^{2}  q^{2}) = q (p + q) (p  q), so 2 = q (p + q) and hence q^{2} + qp + 2 = 0.
Similarly, r^{2} + rp + 2 = 0.
(ii) Line QR has gradient (2aq  2ar) / (aq^{2}  ar^{2}) = 2(q  r) / [ (q  r) (q + p) ] = 2 (p + q)^{ 1}.
r^{2} + rp + 2 = 0 = q^{2} + qp + 2
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 25062016 00:05

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 25062016 00:08
(Original post by riquix)
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y^{2} = 4ax in parametric form is: x = at^{2}, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t^{ 1}.
So the gradient of the normal to the curve at the point with parameter t is t.
Hence, the normal to the curve at Q has gradient q, so has equation
y  2aq = q (x  aq^{2}).
Since this line passes through P, we substitute y = 2ap and x = ap^{2}. Dividing through by a, we have
2 (p  q) = q (p^{2}  q^{2}) = q (p + q) (p  q), so 2 = q (p + q) and hence q^{2} + qp + 2 = 0.
Similarly, r^{2} + rp + 2 = 0.
(ii) Line QR has gradient (2aq  2ar) / (aq^{2}  ar^{2}) = 2(q  r) / [ (q  r) (q + p) ] = 2 (p + q)^{ 1}.
r^{2} + rp + 2 = 0 = q^{2} + qp + 2
=> r^{2}  q^{2} = (r  q) (r + q) = p (q  r)
=> r + q = p
=> gradient of line QR is 2 p^{ 1}
=> equation of QR is y = 2 p^{ 1} (x  ar^{2}) + 2ar.
Let y = 0. Then x = a (r^{2} + rp) = 2a.
So QR passes through ( 2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p^{ 1} x.
So, at T, where OP and QR intersect,
2 p^{ 1} x = 2 p^{ 1} (x  ar^{2}) + 2ar, which implies that
x = 1/2 a (r^{2} + 2rp) = 1/2 a (2) = a.
So T lies on the line x = a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = a into the equation of OP, we find that T has y coordinate 2ap^{ 1}. So the distance of the xaxis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u^{2} + up + 2 = 0, this quadratic must have positive discriminant, i.e. p^{2} > 8, so the modulus of p is greater than 2 (2)^{1/2}.
Hence the modulus of 2ap^{1} is less than a / 2^{1/2}.
^{ }
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 25062016 01:18
(Original post by physicsmaths)
Y coordinate is wrong mate for T. Should 2a/p iirc.
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 25062016 02:40
STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits, a pair of brackets and a power of n missing from the denominator of the first term after "integration by parts" missing but it's still very obvious what should be there  sorry, I did it quite quickly)Last edited by Alex_Aits; 26062016 at 01:09. 
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 25062016 03:40
STEP III Q5
Solution(i)
(ii)
(iv)
Last edited by Farhan.Hanif93; 25062016 at 12:43. Reason: Correction. 
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 25062016 04:04
(Original post by Alex_Aits)
STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits and a pair of brackets missing but it's still very obvious what should be there  sorry, I did it quite quickly)
Attachment 556355556357556350
Attachment 556355556357
Edit: these pictures aren't coming out very clearly at all, but here is q2 and q3
Attachment 556365556367
Seems like my wifi isn't playing ball.. looks like I will have to latex the rest. Although 7,9,10 all need diagrams to do the solution justiceLast edited by Llewellyn; 25062016 at 04:40. 
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 25062016 06:38
STEP III Q6
Latex here doesn't seem to work so i'm uploading a pdf (+latex source +geogebra but I had to add .txt to the filename otherwise TSR doesn't let me upload)
I couldn't do it all correctly. In part (iii), I forgot to consider the case b≥√(1+a²), and in part (iv) I did one of the derivations wrong. How much marks do U think I will lose? 
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 25062016 07:09
STEP III Question 3
Last edited by Mathemagicien; 25062016 at 08:30. 
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 25062016 08:30
STEP III Question 13
Last edited by Mathemagicien; 25062016 at 15:49. 
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 25062016 10:28
(Original post by Mathemagicien)
STEP III Question 3
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 25062016 10:57

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 25062016 11:00
(Original post by Ewanclementson)
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1x)+(1/1x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2x would be the solution as the x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x 
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 25062016 11:13
STEP III Questions 11 and 12
Always a gift or two in the applied sectionLast edited by Mathemagicien; 26062016 at 06:57. 
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 25062016 12:15
(Original post by Mathemagicien)
STEP III Question 13
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 25062016 12:20
(And probably a better method generally, but in this case, calculating E(X^2) was very quick)Last edited by Mathemagicien; 26062016 at 06:58. 
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 25062016 12:32
(Original post by gasfxekl)
For the last part shouldnt you also say that p1,2m+2=p1,2m+1? 
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 25062016 15:42
It was a day filled with mixed emotions. I got 3 full solutions in the first 90 mins then couldn't do much for the last 90 mins.
Score predictions anyone?
Q1  Full
Q2  Did the first part, said that the first part also holds for r and formed equations for OP and QR and attempted to solve for intersection points. Could not proceed through that though.
Q4  Full except for the last part where I did not simplify and let the final answer be 2 (sech y + whatever the ans in 2nd to last part)
Q5  Did the first part, Claimed that (2m+1)! covers Pm+1, 2m+1 and the remaining primes will cancel with m! for the 2nd part. 3rd part I stated that it holds when 2m+1 is composite and couldn't proceed for prime. 4th part I attempted induction with even n (wasn;t successful though) and stated that if it is true for even m then it is true for m+1.
Q8  Full.
Q6  Had 10 minutes left when I began this. Stated the values of R and gamma, and used the formula for arctanh for the conditions for A and B. I then solved for the intersection point, obtaining the result as the instructor said 'STOP WRITING'.
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Updated: July 25, 2016
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